Answer: All pairs $(3n^2, n)$, where $n$ is a positive integer.

Let $k = \frac{3n^2}{m}$ and $\sqrt{n^2+m} = \frac{n\sqrt{k+3}}{\sqrt{k}}$ both should be integers. Since $k$ is integer, $\frac{n\sqrt{k+3}}{\sqrt{k}} \cdot k = n\sqrt{k(k+3)}$ is also integer. Therefore, $k(k+3)$ is a perfect square. If $k > 1$, then since
$$(k+1)^2 = k^2 + 2k + 1 < k(k+3) < k^2 + 4k + 4 = (k+2)^2$$
$k(k+3)$ is not a perfect square. If $k=1$ then $\sqrt{n^2+m} = \sqrt{4n^2} = 2n$ is an integer number. Thus, only solutions are all pairs $(m, n) = (3n^2, n)$, where $n$ is a positive integer.